Question 13 Marks
The diameter of a copper sphere is $18\ cm$. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108m, find its diameter.
AnswerGiven that diameter of a copper sphere $= 18\ cm$
Radius of the sphere $= 9cm$
Length of the wire $= 108m = 10800\ cm$
Volume of cylinder = Volume of sphere
$\pi^2_1\text{h}=\frac{4}{3}\pi\text{r}^3_2$
$\text{r}^2_1\times10800=\frac{4}{3}\times9\times9\times9$
$\text{r}^2_1=0.009$
$\text{r}_1=0.3\text{cm}$
Therefore Diameter $=2\times0.3=0.6\text{cm}$
View full question & answer→Question 23 Marks
How many spherical bullets can be made out of a solid cube of lead whose edge measures $44\ cm$, each bullet being 4cm in diameter?
AnswerIn the given problem, we have a lead cube which is remolded into small spherical bullets Here,
edge of the cube $(s) = 44\ cm$
Diameter of the small spherical bullets $(d) = 4\ cm$ Now,
let us take the number of small bullets be $x$ So,
the total volume of $x$ spherical bullets is equal to the volume of the lead cube.
Therefore, we get, Volume of the $x$ bullets = volume of the cube $\text{x}\Big(\frac{4}{3}\Big)\pi\Big(\frac{\text{d}}{2}\Big)^3=\text{s}^3$
$\text{x}\Big(\frac{4}{3}\Big)\Big(\frac{22}{7}\Big)\Big(\frac{4}{2}\Big)^3=(44)^3$
$\text{x}\Big(\frac{4}{3}\Big)\Big(\frac{22}{7}\Big)(2)^3=85184$
$\text{x}=\frac{(85184)(3)(7)}{(22)(4)(8)}$
$\text{x}=2541$
Therefore, $2541$ small bullets can be made from the given lead cube.
View full question & answer→Question 33 Marks
The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.
AnswerLet the diameter of the earth be $d$ Then,
Diameter of moon will be $\frac{\text{d}}{4}$
Radius of earth $=\frac{\text{d}}{2}$
Radius of moon $=\frac{\frac{\text{d}}{2}}{4}=\frac{\text{d}}{8}$
Surface area of moon $=\frac{4\pi\big(\frac{\text{d}}{8}\big)^2}{4\pi\big(\frac{\text{d}}{2}\big)^2}$
$=\frac{4}{64}=\frac{1}{16}$
Thus the required ratio of the surface areas is $\frac{1}{16}$
View full question & answer→Question 43 Marks
A hemispherical bowl made of brass has inner diameter $10.5\ cm$. Find the cost of tin plating it on the inside at the rate of $Rs.$ $4$ per $100cm^2$
AnswerInner diameter of hemispherical bowl $= 10.5\ cm$ Radius
$=\frac{10.5}{2}=5.25\text{cm}$
Surface area of hemispherical bowl $=2\pi\text{r}^2$
$= 2 \times 3.14 \times (5.25)^2 = 173.25cm^2$
Cost of tin plating $100\ cm^2 area = Rs. 4 $ area
$=\text{Rs}.\ \frac{4\times173.25}{100}=\text{Rs}.6.93$
Cost of tin plating $173.25cm^2$
Thus, the cost of tin plating the inner side of hemispherical bowl is $Rs. 6.93$
View full question & answer→Question 53 Marks
A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is $17.6\ cm$, find the cost of painting it, given the cost of painting is Rs. $5$ per $100cm^2$
AnswerGiven that only the rounded surface of the dome to be painted,
we would need to find the curved surface area of the hemisphere to know the extent of painting that needs to be done.
Now, circumference of the dome $= 17.6\ cm$ Therefore $2\pi\text{r}=17.6$
$2\times\frac{22}{7}\times\text{r}=17.6\text{m}$
So, the radius of the dome $2\pi\text{r}^2$
$=2\times\frac{22}{7}\times2.8\times2.8$
$=49.28\text{m}^2$
Cost of painting $100cm^2$ is $Rs.\ 5$
So, the cost of painting $1m^2 = Rs.\ 500$
Therefore the cost of painting the whole dome
$= Rs.\ 500 \times 49.28 = Rs.\ 24640$
View full question & answer→Question 63 Marks
A cylindrical tub of radius $12\ cm$ contains water to a depth of $20\ cm$. A spherical form ball is dropped into the tub and thus the level of water is raised by $6.75\ cm$. What is the radius of the ball?
AnswerRadius of cylindrical tub $= 12\ cm$
Depth $= 20\ cm$ Let $r$ be the radius of the ball Then Volume of the ball
= Volume of water raised $\frac{4}{3}\pi\text{r}^3=\pi\text{r}^2\text{h}$
$\text{r}^3=\frac{3.14\times(12)^2\times6.75\times3}{4}$
$\text{r}^3=729$
$\text{r}=\sqrt[3]{729}$ $\text{r}=9\text{cm}$
Therefore radius of the ball $= 9\ cm$
View full question & answer→Question 73 Marks
A shopkeeper has one laddoo of radius $5\ cm$. With the same material, how many laddoos of radius $2.5\ cm$ can be made?
AnswerVolume of laddoo having radius $= 5\ cm$ i.e
Volume $(\text{V}_1) =\frac{4}{3\pi\text{r}^3}$ $(\text{V}_1) =\frac{4}{3}\times\frac{22}{7}\times(2.5)^3$
$(\text{V}_1) =\frac{1375}{21}\text{cm}^3$
Therefore number of laddoos $=\frac{\text{V}_1}{\text{V}_1}=\frac{11000}{1375}=8$
View full question & answer→Question 83 Marks
A sphere, a cylinder, and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere. Find the ratio of their volumes.
AnswerLet $r$ be the common radius Height of the cone = height of the cylinder $= 2r$ .
Let $v_1 =$ Volume of sphere $=\frac{4}{3}\pi\text{r}^3$
$v_1 =$ Volume of cylinder
$=\pi\text{r}^2\text{h}=\pi\text{r}^2\times2\text{r}$
$v_1 = $
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}=\frac{1}{3}\pi\text{r}^3$
Now $v_1: v_2: v_3$_$\frac{4}{3}\pi\text{r}^3:2\pi\text{r}^3:\frac{2}{3}\pi\text{r}^3$
$ = 4 : 6 : 2 = 2 : 3 : 1$
View full question & answer→Question 93 Marks
Find the volume of a sphere whose surface area is $154cm2.$
AnswerIn the given problem, we have to find the volume of a sphere whose surface area is given.
So, let us first find the radius of the given sphere.
Surface area of the sphere $(S)$ $= 154cm^2$
Let the radius of the sphere be $r\ cm$
Now, we know that surface area of the sphere
$=4\pi\text{r}^2$
So, $154=4\Big(\frac{22}{7}\Big)(\text{r})^2$
$\text{r}^2=\frac{(154)(7)}{(4)(22)}$
$\text{r}^2=12.25$
Further, solving for r $\text{r}=\sqrt{12.25}$
$\text{r}=3.5$
Therefore, the radius of the given sphere is $3.5\ cm$.
Now, the volume of the sphere $=\frac{4}{3}\pi\text{r}^3$
$=\Big(\frac{4}{3}\Big)\Big(\frac{22}{7}\Big)(3.5)^3$
$=179.66\text{cm}^3$
Therefore, the volume of the given sphere is $179.66\ cm^3.$
View full question & answer→Question 103 Marks
If a sphere is inscribed in a cube, find the ratio of the volume of cube to the volume of the sphere.
AnswerIn the given problem, we are given a sphere inscribed in a cube.
So, here we need to find the ratio between the volume of a cube and volume of sphere.
This means that the diameter of the sphere will be equal to the side of the cube.
Let us take the diameter as $d$.
Here, Volume of a cube $(V_1) = S^3 = d^3$
Volume of a sphere ($V_2$) $=\Big(\frac{4}{3}\Big)\pi\Big(\frac{\text{d}}{2}\Big)^3$
$=\Big(\frac{4}{3}\Big)\pi\Big(\frac{\text{d}^3}{8}\Big)$
$=\frac{\pi\text{d}^3}{6}$
Now, the ratio of the volume of sphere to the volume of the cube
$=\frac{\text{V}_1}{\text{V}_3}$
$\frac{\text{V}_1}{\text{V}_2}=\frac{\text{d}^3}{\Big(\frac{\pi\text{d}^3}{6}\Big)}$
$=\frac{6}{\pi}$
So,the ratio of the volume of cube to the volume of the sphere is $6:\pi.$
View full question & answer→Question 113 Marks
The diameter of a sphere is $6\ cm$. It is melted and drawn into a wire of diameter $0.2\ cm$. Find the length of the wire.
AnswerGiven that Diameter of sphere $= 6\ cm$
Radius of sphere $=\frac{\text{d}}{2}=\frac{6}{2}=3\text{cm}=\text{r}_1$
Diameter of the wire $= 0.2\ cm$
Radius of the wire $= 0.1cm = r_2$
Volume of sphere = Volume of wire
$\frac{4}{3}\pi\text{r}^3_1=\pi\text{r}^2_2\text{h}$
$=\frac{4}{3}\times3\times3\times3=0.1\times0.1\times\text{h}$
$\text{h}=\frac{4\times3\times3}{0.1\times0.1}$
$\text{h}=3600\text{cm}$
$\text{h}=36\text{m}$
Therefore length of wire $= 36\ m$
View full question & answer→Question 123 Marks
If a sphere of radius $2r$ has the same volume as that of a cone with circular base of radius $r$, then find the height of the cone.
AnswerIn the given problem, we are given a cone and a sphere which have equal volumes.
The dimensions of the two are;
Radius of the cone ($r_c$) $= r$
Radius of the sphere ($r_s$) $= 2r$
Now, let the height of the cone $= h$
Here, Volume of the sphere = volume of the cone $\Big(\frac{4}{3}\Big)\pi\text{r}^3_3=\Big(\frac{1}{3}\Big)\pi\text{r}^2_\text{c}\text{h}$
$\Big(\frac{4}{3}\Big)\pi(2\text{r})^3=\Big(\frac{1}{3}\Big)\pi(\text{r})^2\text{h}$
$\Big(\frac{4}{3}\Big)(8\text{r}^3)=\Big(\frac{1}{3}\Big)\text{r}^2\text{h}$
Further, solving for $h$ $\text{h}=\frac{(4)(8\text{r}^3)}{(\text{r}^2)}$
$\text{h}=32\text{r}$
Therefore, the height of the cone is $32r$.
View full question & answer→Question 133 Marks
The diameter of the moon is approximately one-fourth of the diameter of the earth. What is the earth the volume of the moon?
AnswerDiameter of moon $=\frac{1}{4}\text{th}$ diameter of earth Let the diameter of earth be $d$,
so radius $=\frac{\text{d}}{4}$ Radius $=\frac{\frac{\text{d}}{2}}{4}=\frac{\text{d}}{8}$
Volume of moon $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi\Big(\frac{\text{d}}{8}\Big)^3=\frac{4}{3}\times\frac{1}{512}$
Volume of earth $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi\Big(\frac{\text{d}}{2}\Big)^3=\frac{4}{3}\times\frac{1}{8}\pi\text{d}^3$
$\frac{\text{Volume of moon}}{\text{Volume of earth}}=\frac{\frac{4}{3}\times\frac{1}{512}\pi\text{d}^3}{\frac{4}{3}\times\frac{1}{8}\pi\text{d}^3}$
$=\frac{1}{64}$
Thus the volume of the moon is $\frac{1}{64}$ of volume of earth.
View full question & answer→Question 143 Marks
The radius of the internal and external surfaces of a hollow spherical shell are $3\ cm$ and $5\ cm$ respectively. If it is melted and recast into a solid cylinder of height $2\frac{2}{3}$cm. Find the diameter of the cylinder.
AnswerGiven that, Internal radius of the sphere $= 3cm = r_1$
External radius of the sphere $= 5\ cm = r_22$
Height of the cylinder $=\frac{8}{3}\text{cm}=\text{h}$
Volume of the spherical shell = Volume of cylinder
$\frac{4}{3}\pi(\text{r}^3_2-\text{r}^3_1)=\pi\text{r}^2_3\text{h}$
$\frac{4}{3}(5^3-3^3)=\frac{8}{3}\text{r}^2_3$
$\text{r}^2_3=\frac{4\times98\times3}{3\times8}$
$\text{r}_3=\sqrt{49}$
$\text{r}_3=7\text{cm}$
Therefore diameter of the cylinder = $2$(radius) $= 14\ cm$
View full question & answer→Question 153 Marks
A capsule of medicine is in the shape of a sphere of diameter $3.5\ mm$. How much medicine ($mm^3$) is needed to fill this capsule?
AnswerGiven that Diameter of capsule $= 3.5\ mm$ Radius
$\frac{3.5}{2}=1.75\text{mm}$ Volume of spherical sphere
$=\frac{4}{3\pi\text{r}^3}$
$=\frac{4}{3}\times\frac{22}{7}\times(1.75)^3$
$ = 22.458\ mm^3$
Therefore $22.46\ mm^3$ of medicine is required.
View full question & answer→Question 163 Marks
A hemisphere of the lead of radius $7\ cm$ is cast into a right circular cone of height $49\ cm$. Find the radius of the base.
AnswerGiven Radius of the hemisphere = Volume of cone
$\frac{2}{3}\pi\text{r}^3_1=\frac{1}{3}\pi\text{r}^2_2\text{h}$
$\frac{2}{3}\times7^3=\frac{1}{3}\text{r}^2_2\times49$
$\text{r}^2_2=\frac{2\times7\times7\times7\times3}{3\times49}$ $\text{r}^2_2=2058147$
$\text{r}_2=3.47\text{cm}$
Therefore radius of the base $= 3.74\ cm$
View full question & answer→Question 173 Marks
A cylinder of same height and radius is placed on top of a hemisphere. Find the curved surface area of the shape if the length of the shape is $7\ cm$.
AnswerGiven length of the shape $= 7\ cm$
But length $= r + r 2r = 7\ cm$ $r = 3.5\ cm$ Also; $h = r$
Total surface area of shape $=2\pi\text{rh}+2\pi\text{rh}^2$
$=2\pi\text{rr}+2\pi\text{r}^2$
$=2\pi\text{rr}+2\pi\text{r}^2$
$=4\times\frac{22}{7}\times(3.5)^2=154\text{cm}^2$ View full question & answer→Question 183 Marks
A cube of side $4\ cm$ contains a sphere touching its side. Find the volume of the gap in between.
AnswerIt is given that Cube side = 4\ cm
Volume of cube $= (4cm)^3 = 64cm^3$
Diameter of the sphere = Length of the side of the cube $= 4\ cm$
Therefore radius of the sphere $= 2\ cm$
Volume of the sphere
$=\frac{4}{3\pi\text{r}^3}=\frac{4}{3}\times\frac{22}{7}\times(2)^3=33.52\text{cm}^3$
Volume of gap = Volume of cube - Volume of sphere
$= 64cm^3 - 33.52cm^3 = 30.48cm^3$
View full question & answer→Question 193 Marks
The dome of a building is in the form of a hemisphere. Its radius is $63\ dm$. Find the cost of painting it at the rate of Rs. $2$ per sq m.
Answer
Dome radius $ = 63\ dm = 6.3\ m$
Inner surface area of dome $=2\pi\text{r}^2$$ = 2 \times 3.14 \times (6.3)^2 = 249.48m^2$
Now, cost of $1m^2 =$ Rs. $2$
Therefore cost of $249.48m^2 = Rs. (249.48 \times 2) = Rs. 498.96$ View full question & answer→Question 203 Marks
A hemispherical tank is made up of an iron sheet $1\ cm$ thick. If the inner radius is $1\ m$, then find the volume of the iron used to make the tank.
AnswerGiven that, Inner radius of the hemispherical tank$ = 1m = r_1$
Thickness of the hemispherical tank $= 1cm = 0.01m$
Outer radius of hemispherical tank $= (1 + 0.01) = 1.01m = r_2$
Volume of iron used to make the tank $=\frac{2}{3}\pi(\text{r}^3_2-\text{r}^3_1)$
$=\frac{2}{3}\times\frac{22}{7}\big[(1.01)^3-1^3\big]$
$=\frac{44}{21}\big[(1.0303)-1\big]\text{m}^3$ $= 0.06348\ m^3$
View full question & answer→Question 213 Marks
A cylindrical tub of radius 16cm contains water to a depth of $30\ cm$. A spherical iron ball is dropped into the tub and thus level of water is raised by $9\ cm$. What is the radius of the ball?
AnswerLet $r$ be the radius of the iron ball Radius of the cylinder $= 16\ cm$ Then,
Volume of iron ball = Volume of water raised in the hub
$\frac{4}{3}\pi\text{r}^3=\pi\text{r}^2\text{h}$
$\frac{4}{3}\text{r}^3=(16)^3\times9$ $\text{r}^3=\frac{27\times16\times16}{4}$
$\text{r}^3=1728$ $\text{r}=12\text{cm}$
Therefore radius of the ball $= 12\ cm$.
View full question & answer→Question 223 Marks
A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be $1.4\ m$ and its length is $8\ m,$ find the cost of painting it on the outside at the rate of Rs. $10$ per $m^2.$
Answer
Diameter of a cylinder $= 1.4\ m$
Therefore radius of cylinder $= 1.42 = 0.7\ m$
Height of cylinder $= 8\ m$
Therefore surface area of tank $=2\pi\text{rh}+2\pi\text{r}^2$
$=2\times\frac{22}{7}\times0.7\times8+2\times\frac{22}{7}\times(0.7)^2$
$=\frac{176}{5}+\frac{77}{25}=38.28\text{m}^2$
Now cost of $1m^2 = Rs. 10$
Therefore cost of $38.28m^2 = Rs. 382.80$ View full question & answer→